Wavelength, \(\lambda\): distance along the line between two points of a wave which differ in phase by \(360^0\) or \(2\pi\) radian.
If \(\beta\) is the phase shift, then \[\lambda = \dfrac{2\pi}{\beta}\]
The velocity of propagation of wave \[\mbox{velocity} = \dfrac{\lambda}{t}= \lambda \cdot f = \dfrac{2\pi f}{\beta}\]
For the lossless line and \(\beta = 2\pi f \sqrt{LC}\) \[\begin{array}{ccccc} \lambda= & \dfrac{1}{f\sqrt{LC}} & & & \mbox{velocity}=\dfrac{1}{\sqrt{LC}}\end{array}\]
Wavelength of a 50 Hz power transmission \[\lambda \simeq \dfrac{3 \times 10^8}{50} = 6000 ~ \mbox{Km}\]
\[\begin{aligned} \gamma & =\sqrt{yz} =\sqrt{j\omega C\left(r+j\omega L\right)}\\ \alpha+j\beta & =j\omega\sqrt{LC}\left(1-j\dfrac{r}{\omega L}\right)^{1/2}\\ & =j\omega\sqrt{LC}\left(1-j\dfrac{r}{2\omega L}\right) \end{aligned}\] \[\begin{aligned} \alpha & \simeq\dfrac{r}{2}\sqrt{\dfrac{C}{L}}\\ \beta & \simeq\omega\sqrt{LC} \end{aligned}\]
\[\begin{aligned} L & =\dfrac{\mu_{0}}{2\pi}ln\dfrac{D}{r^{'}}\\ C & =\dfrac{2\pi\varepsilon_{0}}{ln\left(D/r\right)}\\ v & =\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}\\ & =\mbox{velocity of light} \end{aligned}\]
\[\begin{aligned} \gamma & =\sqrt{yz}=j\omega\sqrt{LC}\\ \Longrightarrow cosh\left(\gamma l\right) & =cosh\left(j\omega l\sqrt{LC}\right)=cos\left(\omega l\sqrt{LC}\right)\\ \Longrightarrow sinh\left(\gamma l\right) & =sinh\left(j\omega l\sqrt{LC}\right)=jsin\left(\omega l\sqrt{LC}\right) \end{aligned}\]
On simplification, \[\left[\begin{array}{c} V_{S}\\ I_{S} \end{array}\right]=\left[\begin{array}{cc} cos\left(\omega l\sqrt{LC}\right) & jZ_{c}sin\left(\omega l\sqrt{LC}\right)\\ \dfrac{j}{Z_{c}}sin\left(\omega l\sqrt{LC}\right) & cos\left(\omega l\sqrt{LC}\right) \end{array}\right]\left[\begin{array}{c} V_{R}\\ I_{R} \end{array}\right]\] Now if, \(\omega l\sqrt{LC} =n\pi,~n=1,2,3\cdots\) \[\begin{aligned} |V_{S}| & =|V_{R}|\\ |I_{S}| & =|I_{R}| \end{aligned}\]
the receiving-end voltage and current are numerically equal to the corresponding sending-end values,
there is no voltage drop on load
Such a line is called a .
For 50 Hz, the length of line for tuning is \[l=\dfrac{n\pi}{2\pi f\sqrt{LC}}\] Since, \(1/\sqrt{LC}\simeq\vartheta,\) the velocity of light \[\begin{aligned} l & =\dfrac{1}{2}\left(n\lambda\right)=\dfrac{1}{2}\lambda,\lambda,\dfrac{3}{2}\lambda,\cdots\\ & =3000~km,~6000~km,\cdots \end{aligned}\]
too long a distance of transmission from the point of view of cost and efficiency
for a given line, length and frequency tuning can be achieved by increasing \(L\) or \(C\), i.e. by adding series inductances or shunt capacitances at several places along the line length
method is impractical and uneconomical for power frequency lines and is adopted for telephony where higher frequencies are employed
practical method employs series capacitor to reduce the line inductance and shunt inductor to neutralize line capacitance